BASICS OF HYDRAULICS

1) 1)DEFINITIONS 1.1) HYDRAULICS 1.2) CLASSIFICATION 1.2.1) HYDROSTATICS 1.2.2) HYDRODYNAMICS 1.3) FORCE, PRESSURE, AREA 1.4) PASCAL’S LAW 2)MULTIPLICATION OF FORCES 2.1) BRAMAH’S PRESS 2.2) LAW OF CONSERVATION OF ENERGY 3)HYDRAULIC POWER TRANSMISSION 3.1) LINEAR ACTUATOR 3.2) ROTARY ACTUATOR BASIC DEFINITIONS & FORMULAE USES OF HYDRAULICS

BASICS OF HYDRAULICS 4)ADVANTAGES OF HYDRAULICS 4.1) SPEED CONTROL 4.2) DIRECTION CONTROL 4.3) FORCE CONTROL 4.4) OVERLOAD PROTECTION 4.5) COMPACTNESS 5)HOW PRESSURE IS CREATED ADVANTAGES OF HYDRAULICS PRACTICAL DETAILS IN HYDRAULICS

 HYDRAULICS HYDRO AULUS ( meaning Water ) ( meaning Pipe ) HYDRAULICS : Work done by fluids in pipes.

HYDROSTATICS FORCE F 1 AREA A 1 AREA A 2 Eg.:- F 1 = 1 Kg A 1 = 1 Cm 2 P = F 1 = 1 Kg A 1 1 Cm 2 = 1 Kg / Cm 2 ( Same Pressure P ) A 2 = 10 Cm 2 F 2 = P x A 2 = 1 x 10 = 10 Kg HYDROSTATICS FORCE F 2

TURBINE NOZZLE LIQUID AT HIGH VELOCITY HYDRODYNAMICS

IN ORDER TO DETERMINE THE TOTAL FORCE EXERTED ON A SURFACE WE NEED TO KNOW THE PRESSURE OR FORCE PER UNIT AREA. PRESSURE = FORCE FORCE IN  KILOGRAMS ( Kg ) AREA AREA IN  SQ. CM ( Cm 2 ) PRESSURE IN  KILOGRAM / SQ.CM (Kg / Cm 2 ) P = F A FORCE = PRESSURE x AREA THE ATMOSPHERIC AIR EXERTS UNIFORM PRESSURE ALL ROUND. THIS PRESSURE IS APPROX. 1 Kg / Cm 2 AND IS DENOTED AS 1 BAR ( BAROMETER ) PRESSURE F PA

PASCAL’S LAW PRESSURE APPLIED ON A CONFINED FLUID IS TRANSMITTED UNDIMINISHED IN ALL DIRECTIONS AND ACTS WITH EQUAL FORCE ON EQUAL AREAS AND AT RIGHT ANGLES TO THEM. PRESSURE APPLIED ON A CONFINED FLUID IS TRANSMITTED  UNDIMINISHED  IN ALL DIRECTIONS  ACTS WITH EQUAL FORCE ON EQUAL AREAS AND  AT RIGHT ANGLES TO THEM FRENCH SCIENTIST PASCAL DISCOVERED THIS LAW IN THE 17 th CENTURY. RELATES TO USE OF CONFINED FLUID IN  TRANSMITTING POWER  MODIFYING MOTION  MULTIPLYING FORCE.

PASCAL’S LAW FORCE F 1 SMALL AREA A1 FORCE F 2 LARGE AREA A 2 P = F 1 A 1 F 2 = P x A 2 PRESSURE P

BRAMAH’S PRESS HYDRAULIC LEVERAGE MECHANICAL LEVERAGE 10 kg 1Cm 2 10 Cm kg INPUTOUTPUT 10 Kg ON A 1Cm 2 AREA PRESSURE DEVELOPED THROUGHOUT IS 10 Kg / Cm 2 THIS PRESSURE SUPPORTS A WT OF 100 Kg IF AREA IS 10 Cm 2 THE FORCES ARE PROPORTIONAL TO THE PISTON AREAS 10 Kg 1 Cm 2 = 100 Kg 10 Cm 2 A LOAD OF 10 Kg HERE WILL BALANCE A LOAD OF 100 Kg HERE 10 Kg 100 Kg 10 1

1 Cm LAW OF CONSERVATION OF ENERGY 1Cm 2 10 Cm kg 10 kg 10 Cm  ENERGY CAN NEITHER BE CREATED NOR DESTROYED.  WHAT IS GAINED BY FORCE IS SACRIFICED IN THE DISTANCE MOVED. WORK DONE = FORCE x DISTANCE MOVED W = F x d = 10 Kg x 10 Cm = 100 Kg-Cm W = F x d = 100 Kg x 1 Cm = 100 Kg-Cm MOVING THE SMALL PISTON 10 Cm DISPLACES 1 Cm 2 x 10 Cm = 10 Cm 3 OF LIQUID 10 Cm OF LIQUID WILL MOVE LARGER PISTON ONLY 1Cm. 10 Cm 2 x 1 Cm = 10 Cm 3 Q = A x h

HYDRAULIC POWER TRANSMISSION LINEAR ACTUATOR ROTARY ACTUATOR LOAD PUMP PISTON & ROD TO RESERVOIR HYDRO MOTOR

ADVANTAGES OF HYDRAULICS PUMP 10 lpm PISTON MOVES “X” Cm IN 1 min. THIS VOL. IS 10 Lts. MAXIMUM SPEED (No speed control ) (Speed control ) RELIEF VALVE FLOW CONTROL VALVE Q = A x V Q  Flow (Cm 3 /min) A  Area ( Cm 2 ) V  Velocity (Cm/ min ) 5 lpm ACTUATOR GETS ONLY 5 LPM AND TRAVELS “X/2” Cm IN ONE MIN. SPEED CONTROL PUMP

ADVANTAGES OF HYDRAULICS PUMP RELIEF VALVE RELIEF VALVE DIRECTIONAL VALVE DIRECTIONAL VALVE DIRECTION CONTROL THE CYLINDER ROD EXTENDS THE CYLINDER ROD RETRACTS HYDRAULIC DRIVES ARE REVERSIBLE

ADVANTAGES OF HYDRAULICS  RELIEF VALVE PROTECTS THE SYSTEM BY MAINTAINING THE SYSTEM SET PRESSURE.  ANY INCREASE IN PRESSURE IN SYSTEM IS RELEAVED TO TANK. ( MOMENTARILY DIVERTING FLOW TO THE TANK. )  THUS OVERLOAD PROTECTION IS ACHIEVED. OVER LOAD PROTECTION

PRESSURE HEAD PUMP INLET LOCATIONS OIL LEVEL ABOVE PUMP CHARGES INLET 100 Cm PUMP OIL LEVEL BELOW PUMP REQUIRES VACUUM TO “LIFT “ OIL INLET OUTLET PRESSURE HERE IS 0.85 x 100 gm / Cm 2 = Kg / Cm 2 THERE MUST BE A VACUUM EQUIVALENT TO Kg / Cm 2 TO LIFT THE OIL PUMP MECHANISM CREATES THE LOWER PRESSURE CONDITION. 100 Cm INLET OUTLET

HOW PRESSURE IS DEVELOPED NO PRESSURE NO RESTRICTION WITH RESTRICTION CLOSING RELIEF VALVE RELIEF VALVE PRESSURE BUILDS UP PRESSURE BUILDS UPTO RELIEF VALVE SETTING (100 Kg / Cm 2 ) PUMP Set at 100 Kg/Cm 2 RELIEF VALVE PUMP Set at 100 Kg/Cm 2

A B C 10 BAR OPENS VALVE A 20 BAR OPENS VALVE B 30 BAR OPENS VALVE C PARALLEL FLOW PATHS 10 PUMP THE OIL CAN CHOOSE 3 PATHS 20 IF FLOW IS BLOCKED BEYOND “ A” OIL WILL FLOW THRO “B” WHEN PRESSURE REACHES 20 BAR OIL TAKES THE PATH OF LEAST RESISTANCE

SERIES RESISTANCE ADD PRESSURE P 1 = 0 P 2 = ( P ) = = 10 BAR P 3 = ( P ) = = 30 BAR P = ( P ) = = 60 BAR A 10 BAR B 20 BAR C 30 BAR PUMP

PRINCIPLES OF FLOW  HOW FLOW IS MEASURED ? VELOCITY FLOW ( FLOW RATE )  FLOW RATE AND SPEED  FLOW AND PRESSURE DROP  LAMINAR AND TURBULENT FLOW  BERNOULLI’S PRINCIPLE FLOW IS THE ACTION IN THE HYDRAULIC SYSTEM THAT GIVES THE ACTUATOR ITS MOTION. PRESSURE GIVES THE ACTUATOR ITS FORCE, BUT FLOW IS ESSENTIAL TO CAUSE MOVEMENT. FLOW IN THE HYDRAULIC SYSTEM IS CREATED BY THE PUMP PRESSURE INDICATES WORK LOAD.

VELOCITY : IS THE AVERAGE SPEED OF THE FLUID’S PARTICLES PAST A GIVEN POINT OR THE AVERAGE DISTANCE THE PARTICLES TRAVEL PER UNIT OF TIME. Unit :m/Sec or m / min ( Metres / Sec or Metres/min ) FLOW RATE : IS THE VOLUME OF FLUID PASSING A POINT IN A GIVEN TIME. Unit: Cm 3 / min or l / min ( cc / minute or litres / min ) SPEED OF AN ACTUATOR DEPENDES ON THE ACTUATOR SIZE AND RATE OF FLOW INTO IT. Q = A x V FLOW IN Cm 3 / min : AREA IN Cm 2 : VELOCITY IN Cm / min

FLOW AND PRESSURE DROP MAX. PRESSURE HERE BECAUSE OF THE HEAD OF THE FLUID FRICTION IN PIPE DROPS PRESSURE PRESSURE IS ZERO HERE AS THE FLUID FLOWS OUT UNRESTRICTED PRESSURE GRADIENT SUCEEDINGLY LOWER LEVEL OF LIQUID SHOWS PRESSURE IS REDUCED AT POINTS DOWNSTREAM FROM SOURCE. DUE TO EFFECT OF FRICTION RECOMMENDED VELOCITY RANGES ARE : 1.) PUMP INLET LINE  0.6 ~ 1.2 metres / Second 2.) WORKING LINE ( PR. LINES): 2 ~ 6 metres / Second

LAMINAR FLOW TURBULENT FLOW LOW VELOCITY FLOW IN A STRAIGHT PIPE IS STREAMLINED. THE FLUID PARTICLES MOVE PARALLEL TO FLOW DIRECTION. NOR DOES A GRADUAL CHANGE IN DIRECTION. THE FLOW MAY START OUT STREAMLINED. AN ABRUPT CHANGE IN CROSS- SECTION MAKES IT TURBULENT. SO DOES AN ABRUPT CHANGE IN DIRECTION. NON PARALLEL PATHS OF PARTICLES INCREASE RESISTANCE TO FLOW.